NCERT Solutions

we need to determine the magnetic field strength at the center of a multi-turn circular current-carrying coil.

Concept: Biot-Savart Law

The magnetic field at the center of a circular loop is derived from the Biot-Savart Law. For a coil consisting of $N$ closely wound turns, the total magnetic field is the sum of the fields produced by each individual turn. The magnitude of the magnetic field $B$ at the center is given by:

$$B = \frac{\mu_0 N I}{2R}$$

where $\mu_0$ is the permeability of free space, $N$ is the number of turns, $I$ is the current, and $R$ is the radius of the coil.

Given:

• Number of turns: $N = 100$
• Radius of the coil: $R = 8.0\,\text{cm} = 0.08\,\text{m}$
• Current: $I = 0.40\,\text{A}$
• Permeability of free space: $\mu_0 = 4\pi \times 10^{-7}\,\text{T}\cdot\text{m}\cdot\text{A}^{-1}$

Solving:

1. We substitute the given values into the formula for the magnetic field at the center of the coil:

$$\begin{aligned} B &= \frac{(4\pi \times 10^{-7}\,\text{T}\cdot\text{m}\cdot\text{A}^{-1}) \cdot (100) \cdot (0.40\,\text{A})}{2 \cdot (0.08\,\text{m})} \\[10pt] &= \frac{4\pi \times 10^{-5} \cdot 0.40}{0.16}\,\text{T} \\[10pt] &= \frac{1.6\pi \times 10^{-5}}{0.16}\,\text{T} \\[10pt] &= 10\pi \times 10^{-5}\,\text{T} \\[10pt] &= \pi \times 10^{-4}\,\text{T} \end{aligned}$$

2. Using the value of $\pi \approx 3.14159$:

$$\begin{aligned} B &\approx 3.14159 \times 10^{-4}\,\text{T} \\[4pt] &\approx 3.1 \times 10^{-4}\,\text{T} \end{aligned}$$

The magnitude of the magnetic field at the center of the coil is approximately $3.1 \times 10^{-4}\,\text{T}$.

$$3.1 \times 10^{-4}\,\text{T}$$